# How much water does a wave transport into a sea cave?

How much water is transported (volume) as a wave travels into a sea cave?

The wave has a height of 1 m and a period of 12 seconds. The average water depth at the cave mouth it 5 m and the width is 15 m.

How do I calculate this? This is a real-world problem and if there is any other information needed for this problem I will happily collect it.

This isn't a homework problem. This is a real cave and I assumed water is transported because it does pile up in the back of the cave until the wave is reflected and exits the cave in the opposite direction. Just as on beaches where waves push water up on to the beach and gravity pulls the water back to ocean creating the nearshore current. I understand that deep water waves have an almost circular orbit but due to Stokes drift a particle will be slightly displaced in the direction the swell is moving. In this cave we can assume the wave is well with in the shallow water wave criteria and almost at the critical depth in which the wave deformation is extreme and wave energy is about to be converted into turbulent kinetic energy as the wave breaks. The orbit therefore would be linear, as the wave moves in and out. The net change would be 0 m$^3$, but how much water flows in and then flows back out of the cave?

The time-averaged material transport by water waves is quantified by Stokes drift, which is the residual Lagrangian drift due to sub-surface orbits not being closed:

$$u_{St} = \dfrac{\omega k^2 a \cosh[2k(z+d)]}{2\sinh(kd)^2}$$ where $\omega$ is the angular frequency, $k$ is the wavenumber, $a$ is the wave amplitude, $d$ is mean water depth, and $z$ is the displacement from water surface, positive upward.

Because the cave is closed, there is build-up of water in the cave which results in pressure gradient-induced Eulerian backflow which opposes the Stokes drift. This backflow is barotropic in nature and equals:

$$u_E = -\dfrac{1}{d}\int_{-d}^{0}u_{St}(z)dz$$

The transport velocity is the sum of mean Eulerian velocity and Stokes drift:

$$u_L = u_E + u_{St}$$

For the case of $T = 12\ s$, $a = 1\ m$, and $d = 5\ m$, the resulting transport looks like this:

Because the Stokes drift is not uniform in the vertical but compensating Eulerian backflow is, the resulting transport is into the cave in the upper 2 m, and out of the cave below that depth.

• Inviscid ($\mu\nabla^2\mathbf{u}=0$), irrotational ($\nabla \times \mathbf{u} = 0$), and incompressible ($\nabla\cdot \mathbf{u}=0$) flow.
• Small amplitude for the linear wave theory to hold.
• All wave energy is dissipated at the cave wall (no reflection). If there is some reflection of waves out of the cave, then the Stokes drift and the corresponding Eulerian backflow would be reduced in magnitude, but the answer for the net transport would be qualitatively the same.
• Shallow water dispersion relationship ($\omega^2 = gk^2d$).

And here is the Python code:

import numpy as np
import matplotlib.pyplot as plt

T = 12. # wave period [s]
d = 5.  # mean water depth [m]
a = 1.  # wave amplitude [m]
g = 9.8 # gravitational acceleration [m/s]

omega = 2*np.pi/T # angular frequency [rad/s]
k = np.sqrt(omega**2/(g*d)) # wavenumber [rad/m]

z = np.linspace(0,-d,501.) # depth array [m]
dz = -np.gradient(z) # depth increment [m]

# Stokes drift in arbitrary depth
ust = 0.5*omega*k*a**2*np.cosh(2*k*(z+d))/np.sinh(k*d)**2

# Eulerian backflow is the vertical integral of Stokes drift
ue = -np.ones(z.size)*np.sum(ust*dz)/d

fig = plt.figure(figsize=(8,6))